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I read that the Albers projection (https://en.wikipedia.org/wiki/Albers_projection) is area-preserving. The projection is:

$$x=\frac{\sqrt{C-2n\sin \phi}\sin n \lambda}{n}$$ $$y=\rho_0 - \frac{\sqrt{C-2n\sin \phi}\cos n \lambda}{n}$$

So the Jacobean matrix is:

$$\dfrac{d(x, y)}{d(\phi, \lambda)} = \begin{bmatrix} \frac{-\sin n \lambda \cos \phi}{\sqrt{C-2n\sin \phi}} & \frac{\cos n \lambda \cos \phi}{\sqrt{C-2n\sin \phi}} \\ \sqrt{C-2n\sin \phi}\cos n \lambda & \sqrt{C-2n\sin \phi}\sin n \lambda \\ \end{bmatrix}$$

And the Jacobean has determinant

$$-\sin^2 n \lambda \cos \phi - \cos^2 n \lambda \cos \phi = -\cos \phi$$

However, for the projection to be area-preserving, it must have determinant $\pm1$. So where have I gone wrong with my calculations?

Number 34
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    Ah, I think I see it. Is the answer that the projection from latitude/longitude has Jacobean $\cos \phi$? So the two cancel out, making the areas on the Albers projection equal to the corresponding areas on the sphere? – Number 34 Aug 14 '16 at 17:04

1 Answers1

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For the projection to be area-preserving it does not have to have determinant ±1.

See this response to a similar question is a very good explanation https://math.stackexchange.com/q/33184.

Basically,you have to consider the "area element" of the coordinate system. If you compare the jacobean you have calculated to the surface area element in spherical coordinates - you will find they are the same as long as you align your axis. ( Surface Element in Spherical Coordinates contains proofs that the area element is $^2 sin()$ and in our case r=1)

I hope this helps

L_ife
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